Question

Which system of linear equations best represents the tables shown below?

Answer 1

From the first table, we can use the points (-1,3) and (5,-3) to find the first equation:

`\begin{gathered} \text{slope:} \\ m=(y_2-y_1)/(x_2-x_1) \\ \Rightarrow m=(-3-3)/(5-(-1))=(-6)/(5+1)=(-6)/(6)=-1 \\ m=-1 \end{gathered}`

then we can use the point to find the equation in slope-point form:

`\begin{gathered} y-y_1=m(x-x_1) \\ \Rightarrow y-3=-1(x-(-1))=-x-1 \\ \Rightarrow y=-x-1+3=-x+2 \\ y=-x+2 \end{gathered}`

we can do the same with the next table, using the points (1,-1) and (7,11):

`\begin{gathered} \text{slope:} \\ m=(11-(-1))/(7-1)=(11+1)/(6)=(12)/(6)=2 \\ m=2 \\ \text{Equation:} \\ y-(-1)=2(x-1)=2x-2 \\ \Rightarrow y+1=2x-2 \\ \Rightarrow y=2x-2-1=2x-3 \\ y=2x-3 \end{gathered}`

therefore, the system of equations is:

`\begin{cases}y=-x+2 \\ y=2x-3\end{cases}`